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Energy factor of the dehumidifier in IdealLoadsAirSystem

asked 2018-07-09 13:00:47 -0500

MiguelPereira gravatar image

updated 2018-07-09 13:52:31 -0500


I am using IdealLoadsAirSystem to simulate an ideal system with heating, cooling, and dehumidification (latent cooling).

Post-simulation, I can apply the COP and EER of a heat pump to the sensible heat results if I want to estimate more realistic sensible loads. How would I do a similar thing for the latent cooling component, being that a dehumidifier's efficiency is usually rated as liters of water removed per kilowatt-hour of energy consumed (i.e. energy factor)?

Is there a default energy factor value that EnergyPlus uses for the IdealLoadsAirSystem?

Thanks in advance.

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answered 2018-07-09 13:46:20 -0500

Why not just model zonal heat pumps with the ZoneHVAC:PackagedTerminalHeatPump object?

The COP will be dependent on the outdoor air temperature and load, so you can't just take a rated COP and apply it to the total sensible load. You'd need to know the heat pump performance curves and apply them. Doable in a spreadsheet, but EnergyPlus does this calculation for you.

Also, a dehumidifier acts on air within a zone, which will be at a different temperature and humidity level than the input to the heat pump if it includes outdoor air ventilation. The ratings are based on a consistent zone temperature, not on outdoor air as well, so you can't just use them directly on outdoor air. Is the intent to capture latent cooling of just the outdoor air, or outdoor air and space latent loads as well?

Depending on your cooling coil leaving temperature, the coil will provide dehumidification, meaning sensible and latent cooling go together unless less there is a dedicated desiccant-based separate sensible and latent cooling (SSLC) system.

Assuming somehow the system is SSLC with a zone dehumidifier rated in terms of L/kWh, the energy to meet the latent component can be calculated by [latent cooling load (kJ)]* [1 / latent heat of water (kJ/kg)] * [1 / density of water (kg/m3)] * [1000 (L/m3)] * [1 / energy factor(L/kWh)]. Note the latent cooling load here is the zone latent load, not the load from the ideal loads air system (unless there is no outdoor air).

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Thanks for your reply! My problem when trying to solve your expression is that I do not know the energy factor that the IdealLoadsAirSystem uses and which leads to the latent cooling loads that are being reported.

Is there no way of knowing this factor? Or perhaps some variable that reports the amount of water removed from the zone air?

MiguelPereira gravatar imageMiguelPereira ( 2018-07-09 14:29:37 -0500 )edit

There isn't an energy factor; ideals air loads just tells you the idealized latent cooling load. COP = 1. You'll need to calculate the amount of water removal from the latent load.

mdahlhausen gravatar imagemdahlhausen ( 2018-07-09 16:22:33 -0500 )edit

By doing just "[latent cooling load (kJ)]* [1 / latent heat of water (kJ/kg)] * [1 / density of water (kg/m3)] * [1000 (L/m3)]" I arrive at 1624 L of water, given my (yearly total) latent cooling load from IdealLoadsAirSystem. Is this what you meant by "calculate the amount of water removal from the latent load"?

Can I say that "an idealized dehumidifier would remove 1624 L of water with that latent cooling load"?

MiguelPereira gravatar imageMiguelPereira ( 2018-07-11 18:09:20 -0500 )edit

That's correct. Does your system have outdoor air, and how big is the space? That number is low for an ideal air system with outdoor air serving a large space in a humid climate.

mdahlhausen gravatar imagemdahlhausen ( 2018-07-12 10:34:27 -0500 )edit

My total conditioned area is approximately 60 square metres, and I'm only dehumidifying down to 50% relative humidity. The building is in a Mediterranean climate.

MiguelPereira gravatar imageMiguelPereira ( 2018-07-12 17:15:47 -0500 )edit

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Asked: 2018-07-09 13:00:47 -0500

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Last updated: Jul 09